Edit to “Solution”

astrosander edited Dec 23, 2024, 10:46 PM

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revision #12545 parent #11425 ← older

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			### Statement

			$9.2.7.$ Long straight wires with current intersect at an angle $\alpha$. Find the magnetic field induction on a straight line passing through the point of intersection of the wires perpendicular to both of them. The current in the wires is $I$.

			### Solution
		@@ -6,8 +6,8 @@Solution

			![ Direction of magnetic induction created by current in wires \|464x292, 39%](../../img/9.2.7/9.2.7_1.png)

		−	We find the total vector of magnetic induction as the sum of the vectors of magnetic induction created by each of the vectors $$ \vec{B} = \vec{B_1} + \vec{B_2}~~\quad(1)~~ $$ According to the right-hand rule, we can determine the direction of the magnetic induction lines and notice that $\vec{B_1}$ and $\vec{B_2}$ lie in the same plane, parallel to the plane of the wires, at an angle of $\pi - \alpha$. Then, the total magnetic induction vector $\vec{B}$ from the expression $(1)$, then when adding through the cosine theorem will be an adjacent angle — $\alpha$ $$ \boxed{B=\sqrt{B_1^2+B_2^2 - 2B_1B_2\cos\alpha}}~~\quad(2)~~ $$ We will find the magnetic induction of each of them as the induction of an infinite single-horned wire. $$ B_1 = B_2 = \frac{\mu_0 I}{2\pi R}~~\quad(3)~~ $$ We substitute the obtained expression $(3)$ into $(2)$ and obtain the desired magnetic field induction $$ B = \frac{\mu_0 I}{2\pi R} \sqrt{2-2\cos\alpha} \Rightarrow \boxed{B = \frac{\mu_0 I}{\pi R}\sin\left(\frac{\alpha}{2}\right)} $$
		+	We find the total vector of magnetic induction as the sum of the vectors of magnetic induction created by each of the vectors $$ \vec{B} = \vec{B_1} + \vec{B_2}\tag{1} $$ According to the right-hand rule, we can determine the direction of the magnetic induction lines and notice that $\vec{B_1}$ and $\vec{B_2}$ lie in the same plane, parallel to the plane of the wires, at an angle of $\pi - \alpha$. Then, the total magnetic induction vector $\vec{B}$ from the expression $(1)$, then when adding through the cosine theorem will be an adjacent angle — $\alpha$ $$ \boxed{B=\sqrt{B_1^2+B_2^2 - 2B_1B_2\cos\alpha}}\tag{2} $$ We will find the magnetic induction of each of them as the induction of an infinite single-horned wire. $$ B_1 = B_2 = \frac{\mu_0 I}{2\pi R}\tag{3} $$ We substitute the obtained expression $(3)$ into $(2)$ and obtain the desired magnetic field induction $$ B = \frac{\mu_0 I}{2\pi R} \sqrt{2-2\cos\alpha} \Rightarrow \boxed{B = \frac{\mu_0 I}{\pi R}\sin\left(\frac{\alpha}{2}\right)} $$

			#### Answer

			$$B = \frac{\mu_0 I}{\pi R}\sin\left(\frac{\alpha}{2}\right)$$

	### Statement		### Statement

	$9.2.7.$ Long straight wires with current intersect at an angle $\alpha$. Find the magnetic field induction on a straight line passing through the point of intersection of the wires perpendicular to both of them. The current in the wires is $I$.		$9.2.7.$ Long straight wires with current intersect at an angle $\alpha$. Find the magnetic field induction on a straight line passing through the point of intersection of the wires perpendicular to both of them. The current in the wires is $I$.

	### Solution		### Solution
	@@ -6,8 +6,8 @@Solution

	![ Direction of magnetic induction created by current in wires \|464x292, 39%](../../img/9.2.7/9.2.7_1.png)		![ Direction of magnetic induction created by current in wires \|464x292, 39%](../../img/9.2.7/9.2.7_1.png)

	We find the total vector of magnetic induction as the sum of the vectors of magnetic induction created by each of the vectors $$ \vec{B} = \vec{B_1} + \vec{B_2}~~\quad(1)~~ $$ According to the right-hand rule, we can determine the direction of the magnetic induction lines and notice that $\vec{B_1}$ and $\vec{B_2}$ lie in the same plane, parallel to the plane of the wires, at an angle of $\pi - \alpha$. Then, the total magnetic induction vector $\vec{B}$ from the expression $(1)$, then when adding through the cosine theorem will be an adjacent angle — $\alpha$ $$ \boxed{B=\sqrt{B_1^2+B_2^2 - 2B_1B_2\cos\alpha}}~~\quad(2)~~ $$ We will find the magnetic induction of each of them as the induction of an infinite single-horned wire. $$ B_1 = B_2 = \frac{\mu_0 I}{2\pi R}~~\quad(3)~~ $$ We substitute the obtained expression $(3)$ into $(2)$ and obtain the desired magnetic field induction $$ B = \frac{\mu_0 I}{2\pi R} \sqrt{2-2\cos\alpha} \Rightarrow \boxed{B = \frac{\mu_0 I}{\pi R}\sin\left(\frac{\alpha}{2}\right)} $$		We find the total vector of magnetic induction as the sum of the vectors of magnetic induction created by each of the vectors $$ \vec{B} = \vec{B_1} + \vec{B_2}\tag{1} $$ According to the right-hand rule, we can determine the direction of the magnetic induction lines and notice that $\vec{B_1}$ and $\vec{B_2}$ lie in the same plane, parallel to the plane of the wires, at an angle of $\pi - \alpha$. Then, the total magnetic induction vector $\vec{B}$ from the expression $(1)$, then when adding through the cosine theorem will be an adjacent angle — $\alpha$ $$ \boxed{B=\sqrt{B_1^2+B_2^2 - 2B_1B_2\cos\alpha}}\tag{2} $$ We will find the magnetic induction of each of them as the induction of an infinite single-horned wire. $$ B_1 = B_2 = \frac{\mu_0 I}{2\pi R}\tag{3} $$ We substitute the obtained expression $(3)$ into $(2)$ and obtain the desired magnetic field induction $$ B = \frac{\mu_0 I}{2\pi R} \sqrt{2-2\cos\alpha} \Rightarrow \boxed{B = \frac{\mu_0 I}{\pi R}\sin\left(\frac{\alpha}{2}\right)} $$

	#### Answer		#### Answer

	$$B = \frac{\mu_0 I}{\pi R}\sin\left(\frac{\alpha}{2}\right)$$		$$B = \frac{\mu_0 I}{\pi R}\sin\left(\frac{\alpha}{2}\right)$$